Question

A delivery truck travels with a constant velocity up an 8 slope. A 60 kg box sits on the floor of the truck and, because of static friction, does not slide. (a) While the truck travels 300 m up the slope, what is the net work done on the box? (b) What is the work done on the box by the force of gravity? (c) What is the work done on the box by the normal force? (d) What is the work done by friction?

Answer 1

Explanation:

Given that,

The slope of the ramp, $\theta=8^{\circ}$

Mass of the box, m = 60 kg

(a) Distance covered by the truck up the slope, d = 300 m

Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

$W=\dfrac{1}{2}m(v^2-u^2)=0$

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

$W=Fd\ cos\theta$

Here, $\theta=90+8=98^{\circ}$

$W=mgd\ cos\theta$

$W=60\times 9.8\times 300\ cos(98)$

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

(d) The work done by friction is given by :

$W_f=-W$

$W_f=24550.13\ J$

Hence, this is the required solution.